The focal lengths of the objective and eye piece of a microscope are `2 cm and 5 cm` respectively, and the distance between them is `20 cm`. Find the distance of the object from the objective when the final image seen by the eye is `25 cm` from the eye piece. What is the magnifying power ?

Answer

`u_(0) = - 2.3 cm, m = 41.5`

Explanation

Here, `f_(0) = 2 cm, f_(e) = 5 cm, u_(0) = ?, m = ?`
As final image is at `25 cm` from eye piece, threfore, `v_(e) = - 25 cm`.
From `(1)/(u_(e)) = (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(5) = (-6)/(25)`
`u_(e) = -(25)/(6) cm`
Distance of image from objective,
`v_(0) = 20 - (25)/(6) = (95)/(6) cm`
`(1)/(u_(0)) = (1)/(v_(0)) - (1)/(f_(0)) = (6)/(95) - (1)/(2) = -(83)/(2 xx 95)`
`u_(0) = -(2 xx 95)/(83) = - 2.3 cm`
Magnifying power `= m_(0') xx m_(e) = (v_(0))/(u_(0))(1 + (d)/(f_(e)))`
`=(95//6)/(2 xx 95//83)(1 + (25)/(5)) = 41.5`